General Physics (PHY 2130)


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1 General Physics (PHY 130) Lecture 0 Rotational dynamics equilibrium nd Newton s Law for rotational motion rolling Exam II review
2 Lightning Review Last lecture: 1. Momentum: momentum conservation collisions in one and two dimensions Review Problem: An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. She does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. A highway curve has a radius of 85 m. At what angle should the road be banked so that a car traveling at 6.8 m/s has no tendency to skid sideways on the road? (Hint: No tendency to skid means the frictional force is zero.)
3 Review Problem An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to round the curve without skidding. She does so by banking the road in such a way that the force causing the centripetal acceleration will be supplied by the component of the normal force toward the center of the circular path. A highway curve has a radius of 85 m. At what angle should the road be banked so that a car traveling at 6.8 m/s has no tendency to skid sideways on the road? (Hint: No tendency to skid means the frictional force is zero.) y θ N Take the car s motion to be into the page. Draw FBD and apply Newton s Second Law: v ( 1) Fx = N sinθ = mar = m r F = N cosθ w = ( ) 0 y w θ x Divide (3) by (4): Rewrite (1) and (): ( 3) N sinθ = m v r ( 4) N cosθ = mg ( 6.8 m/s) ( 9.8 m/s )( 85 m) v tanθ = = gr θ = 5.1 = 0.089
4 Last time: what if two or more different forces act on lever arm?
5 Net Torque The net torque is the sum of all the torques produced by all the forces Remember to account for the direction of the tendency for rotation Counterclockwise torques are positive Clockwise torques are negative
6 Example 1: Determine the net torque: N 4 m m Given: weights: w 1 = 500 N w = 800 N lever arms: d 1 =4 m d = m 500 N 1. Draw all applicable forces 800 N Find: Στ =?. Consider CCW rotation to be positive τ = (500 N)(4 m) + ( )(800 N)( m) =+ 000 N m 1600 N m =+ 400 N m Rotation would be CCW
7 Where would the 500 N person have to be relative to fulcrum for zero torque?
8 Example : N y d m m Given: weights: w 1 = 500 N w = 800 N lever arms: d 1 =4 m Στ = 0 Find: d =? 500 N 800 N 1. Draw all applicable forces and moment arms τ τ RHS LHS = (800 N)( m) = (500 N)( d m) 800 [ N m] d [ N m] = 0 d = 3. m According to our understanding of torque there would be no rotation and no motion! What does it say about acceleration and force? Thus, according to nd Newton s law ΣF=0 and a=0! F i = ( 500 N) + N' + ( 800 N) = 0 N' = 1300 N
9 Torque and Equilibrium First Condition of Equilibrium The net external force must be zero ΣF = 0 ΣF x = 0 and ΣF y = 0 This is a necessary, but not sufficient, condition to ensure that an object is in complete mechanical equilibrium This is a statement of translational equilibrium Second Condition of Equilibrium The net external torque must be zero Στ = 0 This is a statement of rotational equilibrium
10 Axis of Rotation So far we have chosen obvious axis of rotation If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the net torque The location of the axis of rotation is completely arbitrary Often the nature of the problem will suggest a convenient location for the axis When solving a problem, you must specify an axis of rotation Once you have chosen an axis, you must maintain that choice consistently throughout the problem
11 Center of Gravity (center of mass) The force of gravity acting on an object must be considered In finding the torque produced by the force of gravity, all of the weight of the object can be considered to be concentrated at one point
12 Calculating the Center of Gravity 1. The object is divided up into a large number of very small particles of weight (mg). Each particle will have a set of coordinates indicating its location (x,y) 3. The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm 4. We wish to locate the point of application of the single force, whose magnitude is equal to the weight of the object, and whose effect on the rotation is the same as all the individual particles. This point is called the center of gravity of the object
13 Coordinates of the Center of Gravity The coordinates of the center of gravity can be found from the sum of the torques acting on the individual particles being set equal to the torque produced by the weight of the object x cg = Σm x i Σm i i and y cg = Σm y i Σm i i The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry. Often, the center of gravity of such an object is the geometric center of the object.
14 Example: Find center of gravity of the following system: Given: masses: m 1 = 5.00 kg m =.00 kg m 3 = 4.00 kg lever arms: d 1 =0.500 m d =1.00 m Find: Center of gravity x cg = m x i m i i m1x1 + mx + m3x = m + m + m kg( 0.500m) +.00 kg(0 m) kg(1.00 m) = 11.0 kg = 0.136m 3 3
15 Experimentally Determining the Center of Gravity The wrench is hung freely from two different pivots The intersection of the lines indicates the center of gravity A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object s center of gravity
16 Equilibrium, once again A zero net torque does not mean the absence of rotational motion An object that rotates at uniform angular velocity can be under the influence of a zero net torque This is analogous to the translational situation where a zero net force does not mean the object is not in motion
17 More on Free Body Diagrams Isolate the object to be analyzed Draw the free body diagram for that object Include all the external forces acting on the object
18 Example Suppose that you placed a 10 m ladder (which weights 100 N) against the wall at the angle of 30. What are the forces acting on it and when would it be in equilibrium?
19 Example: Given: weights: w 1 = 100 N length: l=10 m angle: α=30 Στ = 0 Find: f =? n=? P=? 1. Draw all applicable forces. Choose axis of rotation at bottom corner (τ of f and n are 0!) Torques: Forces: L τ = mg cos30 PLsin = 100 N P 1 P = 86.6 N α = 0 F F mg x y = f P = 0 f = 86.6 N = n mg = 0 n = 100 N f 86.6 N Note: f = µ s n, so µ s = = = n 100 N
20 So far: net torque was zero. What if it is not?
21 Torque and Angular Acceleration When a rigid object is subject to a net torque ( 0), it undergoes an angular acceleration The angular acceleration is directly proportional to the net torque The relationship is analogous to F = ma Newton s Second Law
22 Torque and Angular Acceleration F F r, ( ma ) tangential a t t t = = ma = rα, so t multiply by t F t r = r r acceleration : mr α torque τ dependent upon object and axis of rotation. Called moment of inertia I. Units: kg m I Σm r i i τ = Iα The angular acceleration is inversely proportional to the analogy of the mass in a rotating system
23 Newton s Second Law for a Rotating Object The angular acceleration is directly proportional to the net torque The angular acceleration is inversely proportional to the moment of inertia of the object Στ = Iα There is a major difference between moment of inertia and mass: the moment of inertia depends on the quantity of matter and its distribution in the rigid object. The moment of inertia also depends upon the location of the axis of rotation
24 Example: Consider a flywheel (cylinder pulley) of mass M=5 kg and radius R=0. m and weight of 9.8 N hanging from rope wrapped around flywheel. What are forces acting on flywheel and weight? Find acceleration of the weight. mg
25 Example: N I = 1 MR = 0.10 kg m T Given: Mg T masses: M = 5 kg weight: w = 9.8 N radius: R=0. m mg 1. Draw all applicable forces Find: Forces=? a t T = = αr or or I R a t Forces: Torques: F = mg T = ma need T! Tangential acceleration at the edge of flywheel (a=a t ): a t TR = I 0.10 kg m = ( 0. m) a t = (.5kg ) a t a = F = mg T = ma mg mg τ = T R = I α (.5kg ) ( m +.5kg ) a t 9.8 N = 3.5kg T R α = I = ma =.8m s
26 Note on problem solving: The same basic techniques that were used in linear motion can be applied to rotational motion. Analogies: F becomes τ, m becomes I and a becomes α becomes ω and x becomes θ Techniques for conservation of energy are the same as for linear systems, as long as you include the rotational kinetic energy Problems involving angular momentum are essentially the same technique as those with linear momentum The moment of inertia may change, leading to a change in angular momentum, v
27 Review before Exam Useful tips: 1. Do and understand all the homework problems.. Review and understand all the problems done in class. 3. Review and understand all the problems done in the textbook (chapters 47). 4. Come to office hours if you have questions!!!
28 Exam Review
29 Review problem The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown in Figure 1. If the spring is compressed a distance of 0.10 m and the gun fired vertically as shown, the gun can launch a 0.0g projectile from rest to a maximum height of 0.0 m above the starting point of the projectile. Neglecting all resistive forces, determine (a) the spring constant and (b) the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in Figure 1.
30
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